calculate ∫_0 ^∞ e^(−2x) (√(1+e^(−4x) ))dx .


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Question Number 38126 by maxmathsup by imad last updated on 22/Jun/18

$c a l c u l a t e \int_{0}^{\infty} e^{- 2 x} \sqrt{1 + e^{- 4 x}} d x .$
Commented by math khazana by abdo last updated on 26/Jun/18
$let I = ∫_0 ^∞ e^(−2x) (√(1+e^(−4x) ))dx changement e^(−2x) =t give −2x =ln(t)and I = −∫_0 ^1 t(√(1+t^2 ))((−dt)/(2t)) =(1/2) ∫_0 ^1 (√(1+t^2 ))dt 2I =_(t=sh(u)) ∫_0 ^(ln(1 +(√2))) ch(t)ch(t)dt =(1/2) ∫_0 ^(ln(1+(√2))) (1+ch(2t))dt =((ln(1+(√2)))/2) +(1/4) [sh(2t)]_0 ^(ln(1+(√2))) =((ln(1+(√2)))/2) +(1/4)sh(2ln(1+(√2))) =((ln(1+(√2))/2) +(1/8){(1+(√2))^2 −(1+(√2))^(−2) } .$
$l e t I = \int_{0}^{\infty} e^{- 2 x} \sqrt{1 + e^{- 4 x}} d x c h a n g e m e n t$ $e^{- 2 x} = t g i v e - 2 x = l n (t) a n d$ $I = - \int_{0}^{1} t \sqrt{1 + t^{2}} \frac{- d t}{2 t} = \frac{1}{2} \int_{0}^{1} \sqrt{1 + t^{2}} d t$ $2 I =_{t = s h (u)} \int_{0}^{l n (1 + \sqrt{2})} c h (t) c h (t) d t$ $= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} (1 + c h (2 t)) d t$ $= \frac{l n (1 + \sqrt{2})}{2} + \frac{1}{4} {[s h (2 t)]}_{0}^{l n (1 + \sqrt{2})}$ $= \frac{l n (1 + \sqrt{2})}{2} + \frac{1}{4} s h (2 l n (1 + \sqrt{2}))$ $= \frac{l n (1 + \sqrt{2}}{2} + \frac{1}{8} {{(1 + \sqrt{2})}^{2} - {(1 + \sqrt{2})}^{- 2}} .$
Commented by math khazana by abdo last updated on 26/Jun/18
$I =((ln(1+(√2)))/4) +(1/(16)){ (1+(√2))^2 −(1+(√2))^(−2) }.$
$I = \frac{l n (1 + \sqrt{2})}{4} + \frac{1}{16} {{(1 + \sqrt{2})}^{2} - {(1 + \sqrt{2})}^{- 2}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

	$t = e^{- 2 x} d t = - 2 e^{- 2 x} d x$ $= \frac{- 1}{2} \int_{1}^{0} \sqrt{1 + t^{2}} d t$ $= \frac{1}{2} \int_{0}^{1} \sqrt{1 + t^{2}} d t$ $u s e f o r m u l a \int \sqrt{x^{2} + a^{2}} d x$
	Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

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